What is the extreme f(x)= 1/3 x^3-x^2+x+1,0<= x<= 5 ?

The extreme f(x)= 1/3 x^3-x^2+x+1,0<= x<= 5 is Minimum(0,1),Saddle(1, 4/3),Maximum(5, 68/3)

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extreme f(x)= 1/3 x^3-x^2+x+1,0<= x<= 5

extreme points

f (x) = \frac{1}{3} x^{3} - x^{2} + x + 1, 0 \leq x \leq 5

Minimum (0, 1), Saddle (1, \frac{4}{3}), Maximum (5, \frac{68}{3})

Solution steps

Find the critical points:

x = 0, x = 1, x = 5

Domain of

\frac{1}{3} x^{3} - x^{2} + x + 1, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 5

Plug the extreme point

x = 0

into

\frac{1}{3} x^{3} - x^{2} + x + 1 \Rightarrow y = 1

Minimum (0, 1)

Plug the extreme point

x = 1

into

\frac{1}{3} x^{3} - x^{2} + x + 1 \Rightarrow y = \frac{4}{3}

Saddle (1, \frac{4}{3})

Plug the extreme point

x = 5

into

\frac{1}{3} x^{3} - x^{2} + x + 1 \Rightarrow y = \frac{68}{3}

Maximum (5, \frac{68}{3})

Minimum (0, 1), Saddle (1, \frac{4}{3}), Maximum (5, \frac{68}{3})

f (x) = 8 x^{3} - 3 x y + y^{3}

f (x, y) = 3 x^{2} + y^{2} - 3 x

y (x, t) = 7 x - 8 a t (- 4.5)

- x^{2} + 5 x - 2

y = 9 - x

What is the extreme f(x)= 1/3 x^3-x^2+x+1,0<= x<= 5 ?
The extreme f(x)= 1/3 x^3-x^2+x+1,0<= x<= 5 is Minimum(0,1),Saddle(1, 4/3),Maximum(5, 68/3)